# A Review About Upports Mechanics Essay

Structural elements carry their loading to other elements or the ground through connections or supports. In order to be able to analyze a structure it is necessary to be clear about the forces that can be resisted at each support. The actual behaviour of a support or connection can be quite complicated. So much so, that if all of the various conditions were considered, the design of each support would be a terribly lengthy process. And yet, the condition of the supports is very important to the behaviour of the elements which are being supported.

In order to facilitate the analysis of a structure, it is often necessary to idealize the behaviour of a support. This is similar to the massless, frictionless pulley in physics homework problems. Even though these pulleys do not exist, they are useful to enable learning about certain issues. It is important to realize that all of the grpahical representations of supports are idealizations of a real connection. Effort should be made to search out and compare the reality with the grpahical and/or numerical model. It is often very easy to forget that the reality can be strikingly different than what is assumed in the idealization!

The four types of supports that can be found in structures are; roller, frictionless surface, pinned, and fixed. The type of support affects the forces and moments that are used to represent these supports. It is expected that these representative forces and moments, if properly calculated, will bring about equilibrium in the structural element.

There is not a single accepted graphical method to represent each of these support types. However, no matter what the representation looks like, the forces that the type can resist is indeed standardized. Almost all supports can be assigned to one of the four types. The usual methods of graphical designation are:

These supports can be at the ends or at any intermediate points along the structural member. They are used to draw free body diagrams (FBD's) which aid in the analysis of all structural members. Each of the support types is a representation of an actual support.

### ROLLER SUPPORTS

Roller supports are free to rotate and translate along the surface upon which the roller rests. The surface can be horizontal, vertical, or sloped at any angle. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. Roller supports are always found at least one end of long bridges so that forces due to thermal expansion and contraction are minimalized. These supports can also take the form of rubber bearings which are designed to allow a limited amount of lateral movement. A roller support cannot provide any resistance to lateral forces. The representation of a roller support includes one force perpendicular to the surface.

### FRICTIONLESS SUPPORTS

Frictionless surface supports are similar to roller supports. The resulting reaction force is always a single force that is perpendicular to, and away from, the surface. They too are often found as supports for long bridges or roof spans. These are often found supporting large structures in zones of frequent seismic activity. The representation of a frictionless support includes one force perpendicular to the surface.

### PINNED SUPPORTS

A pinned support can resist both vertical and horizontal forces but not a moment. They will allow the structural member to rotate, but not to translate in any direction. Many connections are assumed to be pinned connections even though they might resist a small amount of moment in reality. It is also true that a pinned connection could allow rotation in only one direction; providing resistance to rotation in any other direction. The knee can be idealized as a connection which allows rotation in only one direction and provides resistance to lateral movement. The design of a pinned connection is a good example of the idealization of the reality. A single pinned connection is usually not sufficient to make a structure stable. Another support must be provided at some point to prevent any rotation of the structure. The representation of a pinned support includes both horizontal and vertical forces.

### FIXED SUPPORTS

Fixed supports can resist vertical and horizontal forces as well as a moment. Since they restrain both rotation and translation, they are also known as rigid supports. This means that a structure only needs one fixed support in order to be stable. All three equations of equilibrium can be satisfied. A flagpole set into a concrete base is a good example of this kind of support. The representation of fixed supports always includes two forces (horizontal and vertical) and a moment.

### Free body diagram

Block on a ramp (top) and corresponding free body diagram of just the block (bottom).

A free body diagram is a pictorial representation often used by physicists and engineers to analyze the forces acting on a free body. A free body diagram shows all contact and non-contact forces acting on the body. Drawing such a diagram can aid in solving for the unknown forces or the equations of motion of the body. Creating a free body diagram can make it easier to understand the forces, and moments, in relation to one another and suggest the proper concepts to apply in order to find the solution to a problem. The diagrams are also used as a conceptual device to help identify the internal forces-for example, shear forces and bending moments in beams-which are developed within structures.

### Contents

• Construction
• What is included
• What is excluded
• Assumptions

### Construction

A free body diagram consists primarily of a sketch of the body in question and arrows representing the forces applied to it. The selection of the body to sketch may be the first important decision in the problem solving process. For example, to find the forces on the pivot joint of a simple pair of pliers, it is helpful to draw a free body diagram of just one of the two pieces, not the entire system, replacing the second half with the forces it would apply to the first half.

### What is included

The sketch of the free body need include only as much detail as necessary. Often a simple outline is sufficient. Depending on the analysis to be performed and the model being employed, just a single point may be the most appropriate.

All external contacts, constraints, and body forces are indicated by vector arrows labeled with appropriate descriptions. The arrows show the direction and magnitude of the various forces. To the extent possible or practical, the arrows should indicate the point of application of the force they represent.

Only the forces acting on the object are included. These may include forces such as friction, gravity, normal force, drag, or simply contact force due to pushing. When in a non-inertial reference frame, fictitious forces, such as centrifugal force may be appropriate.

A coordinate system is usually included, according to convenience. This may make defining the vectors simpler when writing the equations of motion. The x direction might be chosen to point down the ramp in an inclined plane problem, for example. In that case the friction force only has an x component, and the normal force only has a y component. The force of gravity will still have components in both the x and y direction: mgsin(Î¸) in the x and mgcos(Î¸) in the y, where Î¸ is the angle between the ramp and the horizontal.

### What is excluded

All external contacts and constraints are left out and replaced with force arrows as described above.

Forces which the free body applies to other objects are not included. For example, if a ball rests on a table, the ball applies a force to the table, and the table applies an equal and opposite force to the ball. The FBD of the ball only includes the force that the table causes on the ball.

Internal forces, forces between various parts that make up the system that is being treated as a single body, are omitted. For example, if an entire truss is being analyzed to find the reaction forces at the supports, the forces between the individual truss members are not included.

Any velocity or acceleration is left out. These may be indicated instead on a companion diagram, called "Kinetic diagrams", "Inertial response diagrams", or the equivalent, depending on the author.

### Assumptions

The free body diagram reflects the assumption and simplifications made in order to analyze the system. If the body in question is a satellite in orbit for example, and all that is required is to find its velocity, then a single point may be the best representation. On the other hand, the brake dive of a motorcycle cannot be found from a single point, and a sketch with finite dimensions is required.

Force vectors must be carefully located and labeled to avoid assumptions that presuppose a result. For example, in the accompanying diagram of a block on a ramp, the exact location of the resulting normal force of the ramp on the block can only be found after analyzing the motion or by assuming equilibrium.

Other simplifying assumptions that may be considered include two-force members and three-force members.

### Free-Body Diagram

A free-body diagram is a sketch of an object of interest with all the surrounding objects stripped away and all of the forces acting on the body shown. The drawing of a free-body diagram is an important step in the solving of mechanics problems since it helps to visualize all the forces acting on a single object. The net external force acting on the object must be obtained in order to apply Newton's Second Law to the motion of the object.

 A free-body diagram or isolated-body diagram is useful in problems involving equilibrium of forces. Free-body diagrams are useful for setting up standard mechanics problems.

### Concept of System and Surrounding

The body under observation is known as system and rest of universe as surroundings For example:

If we are observing the body 2 then system will be body 2 only, the rest of universe i.e., block one, earth or the horizontal surface will behave like surroundings.

On the system under consideration we will observe only the force exerted by the surrounding on the system.

### Steps for making Free Body Diagram.

Step 1: Identify the object or system and isolate it from other objects clearly specify its boundary.

Step 2: First draw non contact external force in the diagram. Generally it is weight.

Step 3: Draw contact forces which acts at the boundary of the object or system. Contact forces are normal reaction, friction, tension and applied force.

In a Free Body Diagram, internal forces are not drawn, only external forces are drawn.

Solved Example 1: Draw the free body diagram of block 1.

### Solution:

(In this case the block is one system)

### Free body diagram

A free body diagram is an extremely useful tool for assessing the interaction of forces on bodies This is essentially a sketch of a body which is in equilibrium and is entirely separate from the surroundings.The only rule for drawing free-body diagrams is to depict all the forces which exist for that object in the given situation. Below is shown a typical 2- dimensional free body diagram of a cantilever beam

The free body diagram includes external forces applied to the body and external reaction forces resulting from the method of supporting the body. Some reactions are shown in the 2-dimensional figure below.

### Force

A force can be represented by a localized vector defined by magnitude, direction and point of application. A number of forces applied to any point can be replaced by a single resultant force using the principles of vector addition as shown on the vector page of this website. From Newton's first law it is known that a particle will remain at rest if all the resultant force on the particle = zero. This called the equilibrium condition. Using Cartesian co-ordinate system this may be states as follow;

Or

This is simplified by considering only the scalar identities..

Since there are three equations of equilibrium for the three dimensional case, there can be at the most three unknowns which can be determined from these equations.

### Moments

A moment applied to a body creates a tendency for the body to rotate.. The moment of a force about a point equals the product of a distance (say r) = the lever arm and the force (F) acting perpendicular to the lever arm..

The moment of a force is defined in vector form using determinant algebra (in 2 dimensional and 3 dimensional cases) as

If a number of concurrent forces are applied to a point P and r is the position vector from 0 to P. the moment caused by these forces =

Mo = r x (F1 + F2 + F3...) = r x F1 + r x F2 + r x F3...

That is the moment due to several concurrent forces is equal to the sum of the moments of the individual forces.

### Couples

If two forces are equal in magnitude (F), parallel in line of action on opposite in direction they result in a couple. The magnitude of the moment of the couple is the distance between the force (e) multiplied by one force. The direction of the couple is identified by the right hand rule....

Couple = e x F

It can be proved that the moment of the couple is the same magnitude at any location...

If a number of forces are applied to a body resulting in a number of moments the moments can be combined algebraically to a resultant moment with a force.. An example below shows how two offset forces and a single moment can be combined to a resultant force and moment..

### Rigid Body Equilibrium

A particle is in equilibrium if the resultant force acting upon it is zero... A rigid body is in equilibrium if the resultant force is zero and if the resultant moment = 0...

There are therefore at the most 6 independent equations of equilibrium. The six equations are obtained from the free body diagram showing all of the applied forces and moments and also all of the resulting reaction forces and moments. From these six equations it is possible to solve for six unknowns.If there are more than six unknowns then the system is statically indeterminate...

For the special case of 2 dimensional equilibrium which is most often applied for beams and simple structures the equilibrium equation as below is used...

A number of support systems with the resulting reactions are shown below...

### Structures

Any assemblage of materials whose function is that of supporting loads is a structure.. The term may be applied to a bridge, and aeroplane wing a building or a dam. The component parts of a loaded structure are in a state of stress and the laws which govern the distribution of the stress are used to calculate the design a material to enable the structure to safely support the loads.

Structures are classified into two general groups : framed structures and mass structures. The former is based on a number of separate bars or plates pinned, rivetted or welded together as a lattice. These depend upon the geometric properties of the arrangement to withstand the load. Mass structures depend upon the mass of material in the structure to withstand the resistance to the load e.g a masonary damn. The notes following relate only to framed structures.

The notes that follow relate to frameworks or trusses. These are arrangements of bars connected at pin joints which do not transmit moments.The connecting bars are only allowed to transmit tensile forces (Ties) or compressive forces (struts).It is assumed that the struts and ties experience virtually zero deformation.

For a plane frame (2- dimensional) the number of bars (N) required with J joints.

N = 2.J -3

For a space frame (3- dimensional) the number of bars (N) required with J joints.

N = 3.J - 6

There are three methods of assessing frameworks

 Method of Joints Graphical Method Â Method of Sections

### Method of Joints

The steps in this procedure are listed as follows.

 Label all pin joints - A, B, c etc. Draw and label a free body diagram for the whole framework. Calculate the reaction forces using the three equations for static equilibrium applied to the whole framework Draw a free body diagram for each pin assembly. This may be one free body diagram showing all the pin details Â Apply the two equations of static equilibrium (related to Fx and Fy - no moments at joints) at each joint to identify the forces in the attached bars.

0000000000000

### Graphical Method

Note: this method is convenient for simple frames but is very complicated for larger frames and cannot be used for space frames (3 -dimensions).

The steps in this procedure are listed as follows.

 Label all pin joints - A, B, c etc. Draw and label a free body diagram for the whole framework. Calculate the reaction forces using the three equations for static equilibrium applied to the whole framework Draw a free body diagram for each pin assembly. This may be one free body diagram showing all the pin details Produce a force vector polygon for each joint starting with a joint which has no more than two unknowns Â The force polygons can be combined into a single diagram called a Maxwell diagram
1. An simple example is provided below..

The steps 1, 2 & 3 above have been omitted as they are relatively obvious

1. below includes a free body diagram for the each of the joints
2. below is a polygon of forces for joint A. The forces in the bar directions are drawn to scale.
3. below is a polygon of forces for joint B. the forces in the bar directions are drawn to scale.
4. Below is a polygon of forces for joint C. The forces in the bar directions are drawn to scale.
5. Below is a combined force diagram (Maxwell diagram) for the truss..

### Method of Sections

Note: This method can be used to determine the forces in selected bars more rapidly then with the method of joints in some cases..

The steps in this procedure are listed as follows.

 Label all pin joints - A, B , c etc. Draw and label a free body diagram for the whole framework. Calculate the reaction forces using the three equations for static equilibrium applied to the whole framework Draw a free body diagram for a selected part of the frame which can include two joints and breaks the bars under consideration Â Calculate the forces in the bars to one side of the cut using the 3- equations for static equilibrium

An example illustrating the method of sections is shown below; the object is to determine the forces in members AB and JI

A section line is drawn through the members under consideration and a free body diagram is drawn for the part of the structure to the left of the section line...

The unknown forces are shown in tension.

Taking moments about joint A eliminates Forces FAI and FAB leaving only the force FJI.

Sum Moments about A to zero.

0 = 100.(1,5) - FJI.(1.5) therefore FJI = +100kN (tension)

Taking moments about joint I outside of the free body, eliminates FAI and FJI leaving only FAB

Sum Moments about I to zero.

0 = 50.(1,5) - 14.(1,8) +33,67.(1,8) + FAB therefore FAB. (1,5) = -73,60kN (tension)

Summing moments vertically to zero eliminates both FJI and Force FAB leaving only FAI

0 = 33,67 -14 - FAI (5 / 7.81) therefore FAI = +30,72 (tension)

FJA = -33.67kN (compression) by expansion